The bombardment of Lithium with protons gives rise to the following reaction: `._3Li^7+._1H^1 to .2_He^4+.2_He^4+Q` The atomic masses of lithium, hydrogen and helium are : `7.016u, 1.008u` and `4.004u` resp. Find the initial energy of each of `alpha` particle. Take `1a.m.u=931MeV`.
Text Solution
AI Generated Solution
To solve the problem, we need to find the initial energy of each alpha particle produced in the reaction of lithium with protons. We will follow these steps:
### Step 1: Write down the reaction
The reaction given is:
\[
_3Li^7 + _1H^1 \rightarrow 2_2He^4 + Q
\]
Where \( Q \) represents the energy released in the reaction.
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